- 分类：Tree
- 时间复杂度: O(n)
- 空间复杂度: O(h)
116. Populating Next Right Pointers in Each Node
You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:
struct Node {
int val;
Node *left;
Node *right;
Node *next;
}
Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.
Initially, all next pointers are set to NULL.
Example:
image
Input: {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":{"$id":"6","left":null,"next":null,"right":null,"val":6},"next":null,"right":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}
Output: {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":{"$id":"6","left":null,"next":null,"right":null,"val":7},"right":null,"val":6},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"7","left":{"$ref":"5"},"next":null,"right":{"$ref":"6"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"7"},"val":1}
Explanation: Given the above perfect binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B.
Note:
-
You may only use constant extra space.
-
Recursive approach is fine, implicit stack space does not count as extra space for this problem.
代码：
我自己的代码思路：
"""
# Definition for a Node.
class Node:
def __init__(self, val, left, right, next):
self.val = val
self.left = left
self.right = right
self.next = next
"""
class Solution:
prev_nodes=[]
def connect(self, root: 'Node') -> 'Node':
if not root:
return None
return self.helper(root,0)
def helper(self,root,level):
if root==None:
if self.prev_nodes==[]:
self.prev_nodes=[None] * level
return
self.helper(root.right,level+1)
self.helper(root.left,level+1)
root.next=self.prev_nodes[level]
self.prev_nodes[level]=root
return root
youtube上猫打球的思路代码：
"""
# Definition for a Node.
class Node:
def __init__(self, val, left, right, next):
self.val = val
self.left = left
self.right = right
self.next = next
"""
class Solution:
prev_nodes=[]
def connect(self, root: 'Node') -> 'Node':
if not root:
return None
return self.helper(root,0)
def helper(self,root,level):
if root==None:
if self.prev_nodes==[]:
self.prev_nodes=[None] * level
return
self.helper(root.right,level+1)
self.helper(root.left,level+1)
root.next=self.prev_nodes[level]
self.prev_nodes[level]=root
return root
GTH思路代码的变种：
"""
# Definition for a Node.
class Node:
def __init__(self, val, left, right, next):
self.val = val
self.left = left
self.right = right
self.next = next
"""
class Solution:
def connect(self, root: 'Node') -> 'Node':
if not root:
return None
res=Node(0,None,None,None)
res.left=root
while root!=None:
self.helper(root)
root=root.left
return res.left
def helper(self,root):
if root.left==None and root.right==None:
return
root.left.next=root.right
if root.next==None:
root.right.next==None
else:
root.right.next=root.next.left
if root.next!=None:
self.helper(root.next)
else:
return
讨论：
1.自己写了一个代码，但是为啥这么慢？
2.youtube上的小姐姐讲的很不错，但是为啥还是这么慢？
3.用GTH的思路代码为什么还是这么慢？有毒吧
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